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ZIMSEC A Level · N2010

Chemistry Paper 2 November 2010

Questions
15
Total marks
21
Time allowed
75 min

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Questions
15
Pass mark
9
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]chemical equilibrium
The iodine liberated in an experiment required 15.0 cm315.0\ cm^3 of 0.03 mol dm−30.03\ mol\,dm^{-3} sodium thiosulphate. The number of moles of thiosulphate used is
  1. A4.50×10−44.50 \times 10^{-4}
  2. B4.50×10−34.50 \times 10^{-3}
  3. C4.50×10−24.50 \times 10^{-2}
  4. D2.25×10−42.25 \times 10^{-4}

Question 102

[1 marks]chemical equilibrium
4.5×10−44.5 \times 10^{-4} mol of thiosulphate reacted with the iodine according to I2+2S2O32−→2I−+S4O62−I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}. The number of moles of iodine present is
  1. A1.13×10−41.13 \times 10^{-4}
  2. B2.25×10−42.25 \times 10^{-4}
  3. C4.50×10−44.50 \times 10^{-4}
  4. D9.00×10−49.00 \times 10^{-4}

Question 103

[1 marks]chemical equilibrium
The indicator used in the titration of iodine against sodium thiosulphate, and the colour change at the end point, are
  1. Astarch; colourless to blue-black
  2. Bstarch; blue-black to colourless
  3. Cmethyl orange; red to yellow
  4. Dphenolphthalein; pink to colourless

Question 104

[2 marks]chemical equilibrium
During the titration of the iodine formed in this experiment against sodium thiosulphate, iodine reacts with thiosulphate ions. Write the balanced ionic equation for this reaction, including charges.

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Question 105

[2 marks]chemical equilibrium
A 0.23 g sample of hydrogen iodide, HI (MrM_r = 128), was heated at 800 K in a 200 cm3cm^3 bulb until equilibrium was reached in 2HI⇌H2+I22HI \rightleftharpoons H_2 + I_2. Given that 2.25×10−42.25 \times 10^{-4} mol of I2I_2 was present at equilibrium, calculate the number of moles of HI present at equilibrium.

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Question 106

[2 marks]chemical equilibrium
A 0.23 g sample of HI (MrM_r = 128) was heated at 800 K in a 200 cm3cm^3 bulb until equilibrium was reached in 2HI⇌H2+I22HI \rightleftharpoons H_2 + I_2, giving 2.25×10−42.25\times10^{-4} mol of I2I_2 (and an equal amount of H2H_2) and 1.35×10−31.35\times10^{-3} mol of HI at equilibrium. The value of KcK_c at this temperature is
  1. A0.016
  2. B0.020
  3. C0.028
  4. D0.056

Question 107

[1 marks]chemical equilibrium
In this experiment, the equilibrium mixture at 800 K is suddenly cooled to room temperature before it is titrated. This sudden cooling is done because
  1. Ait shifts the equilibrium position further towards iodine and hydrogen, so more iodine is available to titrate than was actually present at 800 K.
  2. Bit drives the reverse reaction to completion, converting the remaining hydrogen iodide back into hydrogen and iodine before titration.
  3. Cit increases the value of KcK_c at room temperature, so a larger amount of iodine can be detected by the thiosulphate titration.
  4. Dit slows the reaction rate so much that the 800 K equilibrium composition is effectively frozen before titration.

Question 201

[1 marks]group II chemistry
Beryllium forms predominantly covalent compounds, unlike the other Group II elements, because the Be2+Be^{2+} ion
  1. Ahas an incomplete d subshell
  2. Bis easily reduced to the metal
  3. Cis very small and has a high polarising power
  4. Dhas a very low charge density

Question 202

[1 marks]group II chemistry
Barium sulphate is used as a 'barium meal' in the X-ray examination of the gut because it is
  1. Astrongly alkaline
  2. Ba good conductor of electricity
  3. Creadily soluble in water
  4. Dinsoluble and therefore non-toxic

Question 203

[1 marks]group II chemistry
The solubility product of barium hydroxide is 1.35×10−21.35 \times 10^{-2}. Its solubility, in mol dm−3mol\,dm^{-3}, is
  1. A0.058
  2. B0.15
  3. C0.30
  4. D0.60

Question 204

[2 marks]group II chemistry
Down Group (II) from magnesium to barium, the solubility of the metal sulphates decreases mainly because
  1. Athe lattice energy decreases faster than the hydration energy, making the lattice easier to break apart.
  2. Bboth the hydration and lattice energies decrease at the same rate down the group, leaving solubility unchanged.
  3. Cthe hydration energy decreases faster than the lattice energy, making dissolving less energetically favourable.
  4. Dthe sulphate ion becomes more polarising as the cation increases in size, strengthening the lattice.

Question 205

[1 marks]group II chemistry
Epsom salt, MgSO4⋅7H2OMgSO_4 \cdot 7H_2O, is used medicinally as a

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Question 206

[1 marks]group II chemistry
Calcium sulphate, in the form of plaster of Paris, is used to make

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Question 207

[2 marks]group II chemistry
A base is classified as an alkali only if a solution of at least 1×10−1 mol dm−31\times10^{-1}\ mol\,dm^{-3} of it can be made. Barium hydroxide has a solubility of 0.15 mol dm−3mol\,dm^{-3} and strontium hydroxide has a solubility of about 0.034 mol dm−3mol\,dm^{-3}. Which of these two hydroxides qualifies as an alkali?
  1. ABarium hydroxide only: its 0.15 mol dm-3 solubility clears the 0.1 mol dm-3 threshold, unlike strontium hydroxide's 0.034 mol dm-3.
  2. BStrontium hydroxide only: its lower solubility means a smaller, more concentrated sample is needed to reach saturation.
  3. CBoth hydroxides qualify: each forms a saturated solution well above the 0.1 mol dm-3 alkali threshold at room temperature.
  4. DNeither hydroxide qualifies: both solubility values given were measured under different conditions than the titration.

Question 208

[2 marks]group II chemistry
Strontium hydroxide has a solubility product Ksp=4s3=1.57×10−4K_{sp} = 4s^3 = 1.57 \times 10^{-4}. Calculate its solubility, ss, in mol dm−3mol\,dm^{-3}.

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